If $$\int \sqrt{x-\frac{1}{x}}\left(\frac{x^2+1}{x^2}\right) d x=\frac{2}{3}\left(x-\frac{1}{x}\right)^k+c$$, then value of $$k$$ is

$$\int \cot x \cdot \log [\log (\sin x)] d x=$$

$$\int \log x \cdot[\log (e x)]^{-2} d x=\ldots$$

If $\int \frac{1}{1-\cot x} d x=A x+B \log |\sin x-\cos x|+c$ then $A+B=\ldots \ldots$